Question

When 21.7 mL of 0.500 M H2SO4 is added to 21.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Q = 1.387kJ. Calculate ΔH for water formed from this reaction.

Answer 1

H = -63.9 kJ/mol

Explanation

molarity H₂SO₄ = 0.500 M

volume H₂SO₄ solution = 21.7 mL

moles H₂SO₄ = (molarity H₂SO₄) * (volume H₂SO₄ solution in Liter)

moles H₂SO₄ = (0.500 M) * (21.7 x 10^-3 L)

moles H₂SO₄ = 0.01085 mol

moles KOH = (molarity KOH) * (volume KOH in Liter)

moles KOH = (1.00 M) * (21.7 x 10^-3 L)

moles KOH = 0.0217 mol

H₂SO₄ (aq) + 2 KOH (aq) K₂SO₄ (aq) + 2 H₂O (l)

moles H₂O formed = 2 * (moles H₂SO₄) = (moles KOH) = 0.0217 mol

Heat change of solution = 1.387 kJ

Heat change of reaction = -(Heat change of solution)

Heat change of reaction = -(1.387 kJ)

Heat change of reaction = -1.387 kJ

H = (Heat change of reaction) / (moles H₂O formed)

H = (-1.387 kJ) / (0.0217mol)

H = -63.9 kJ/mol